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natta225 [31]
3 years ago
9

A vehicle purchased for $ 27500 depreciates at a constant rate of 4 % . Determine the approximate value of the vehicle 12 years

after purchase.
Mathematics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

$13,200

Step-by-step explanation:

Multiply the original cost by .04 and 12 years

(original cost)(rate in decimal form)(time)

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Marian is making accessories for the soccer team. She uses 666.12 inches of fabric on headbands for 36 players and 3 coaches. Sh
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Answer:

answe is about 23. something

Step-by-step explanation:

5 0
2 years ago
A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions
faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3><u>Solution:</u></h3>

Given that  

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

<em><u>The area of rectangle is given as:</u></em>

\text {Area of rectangle }=length \times width

Area of rectangular parking lot = length of rectangular parking plot \times width of the rectangular parking

\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}

But it is given that Area of rectangular parking lot = 15000 square feet

\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}

Solving the above quadratic equation using quadratic formula

<em><u>General form of quadratic equation is  </u></em>

{ax^{2}+\mathrm{b} x+\mathrm{c}=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}

\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet  

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

3 0
2 years ago
Owen’s bedroom has a perimeter of 46 feet. If the length of the bedroom is 11 feet, what is the width
AnnyKZ [126]

Perimeter of a rectangle is given by (2 × Length(l)) + (2 × Width(w)) = 2l + 2w

l= 11 feet

2(11) + 2w = 46

2w = 46 - 22 = 24

w = 24/2 = 12 feet

Width = 12 feet.

4 0
3 years ago
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2 years ago
(16 divided by 4) x (7-4)
zavuch27 [327]
4 x 3 = 12
12 would be the answer
4 0
3 years ago
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